Question

If the shortest wavelength of hydrogen atom in Lyman series is $x$, then longest wavelength in Paschen series of $He ^{+}$ is

• A. $\frac{9 x}{5}$
• B. $\frac{36 x}{5}$
• C. $\frac{36 x}{7}$
• D. $\frac{5 x}{9}$

Answer 1

Answer: (C)

For shortest wavelength line (maximum energy), $n_{1}=1, n_{2}=\infty, Z=1$

$\frac{1}{\lambda_{\min }}=R \cdot Z^{2}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$

$\Rightarrow \frac{1}{x}=R .1^{2}\left[\frac{1}{1^{2}}-\frac{1}{\infty^{2}}\right]=R$

For Paschen series line having longest wavelength (minimum energy $), n_{1}=3, n_{2}=4$

For $He ^{+}$

$\frac{1}{\lambda_{\max }}=R \times 4\left[\frac{1}{3^{2}}-\frac{1}{4^{2}}\right]$

$\frac{1}{\lambda_{\max }}=\frac{1}{x} \cdot 4\left[\frac{1}{9}-\frac{1}{16}\right]$

$=\frac{1}{x} \times 4 \times\left[\frac{16-9}{9 \times 16}\right]$

$=\frac{7 \times 4}{16 \times 9 \times x}$

$\frac{1}{\lambda_{\max }}=\frac{7}{36 x}$

$\Rightarrow \lambda_{\max .}=\frac{36 x}{7}$